By Daniel Fleisch

ISBN-10: 0511393083

ISBN-13: 9780511393082

Gauss's legislation for electrical fields, Gauss's legislation for magnetic fields, Faraday's legislation, and the Ampere-Maxwell legislations are 4 of the main influential equations in technological know-how. during this consultant for college students, each one equation is the topic of a complete bankruptcy, with targeted, plain-language motives of the actual which means of every image within the equation, for either the fundamental and differential kinds. the ultimate bankruptcy indicates how Maxwell's equations could be mixed to provide the wave equation, the root for the electromagnetic conception of sunshine. This ebook is an excellent source for undergraduate and graduate classes in electromagnetism and electromagnetics. an internet site hosted through the writer at www.cambridge.org/9780521701471 includes interactive ideas to each challenge within the textual content in addition to audio podcasts to stroll scholars via every one bankruptcy.

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**Example text**

In the region x > 3 m, q @ @ @ ~ ~ r E ¼ ¼ ^i þ ^j þ ^k ðb^iÞ ¼ 0; e0 @x @y @k so q ¼ 0 at x ¼ 5 m. Problems The following problems will test your understanding of Gauss’s law for electric fields. Full solutions are available on the book’s website. 1 Find the electric flux through the surface of a sphere containing 15 protons and 10 electrons. Does the size of the sphere matter? 2 A cube of side L contains a flat plate with variable surface charge density of r ¼ À3xy. If the plate extends from x ¼ 0 to x ¼ L and from y ¼ 0 to y ¼ L, what is the total electric flux through the walls of the cube?

2: Given the flux through a closed surface, find the enclosed charge. Problem: A line charge with linear charge density k ¼ 10À12 C/m passes through the center of a sphere. 13 · 10À3 Vm, what is the radius R of the sphere? Charged line L Sphere encloses portion of line Solution: The charge on a line charge of length L is given by q ¼ kL. Thus, UE ¼ qenc kL ¼ ; e0 e0 and L¼ U E e0 : k Since L is twice the radius of the sphere, this means 2R ¼ U E e0 k or R ¼ UE e0 : 2k Inserting the values for UE, e0 and k, you will find that R ¼ 5 · 10À3 m.

To find the flux through the flat surface, use the expression for flux Z UB ¼ ~ B^ n da: S ^ so In this case, ^ n ¼ À’, l I l I 0 ^ ðÀ’Þ ^ ¼À 0 : ’ 2pr 2pr To integrate over the flat face of the half-cylinder, notice that the face lies in the yz plane, and an element of surface area is therefore da ¼ dy dz. 2 The differential form of Gauss’s law The continuous nature of magnetic field lines makes the differential form of Gauss’s law for magnetic fields quite simple. The differential form is written as ~ ~ r B ¼ 0 Gauss’s law for magnetic fields ðdifferential formÞ: The left side of this equation is a mathematical description of the divergence of the magnetic field – the tendency of the magnetic field to ‘‘flow’’ more strongly away from a point than toward it – while the right side is simply zero.

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